Question

A particle is dropped from top of tower of 40 m and simultaneously another particle is projected from ground in upward direction with 20 m/s! Find when and at what distance from ground they will meet

Answer 1

They will meet after 2 seconds at a distance of 20 meters from the ground.

Answer 2

Let the height of the tower be $H$. Let the particle dropped from the top be particle 1 and the particle projected from the ground be particle 2. Let $t$ be the time when they meet and $y$ be the distance from the ground where they meet. We take the origin at the ground and the upward direction as positive.

The initial position of particle 1 is $H$ and its initial velocity is $u_1 = 0$. Its position at time $t$ is given by $y_1(t) = H + u_1 t - \frac{1}{2}gt^2 = H - \frac{1}{2}gt^2$.

The initial position of particle 2 is $0$ and its initial velocity is $u_2$. Its position at time $t$ is given by $y_2(t) = 0 + u_2 t - \frac{1}{2}gt^2 = u_2t - \frac{1}{2}gt^2$.

When the particles meet, their positions are the same, i.e., $y_1(t) = y_2(t)$.

$H - \frac{1}{2}gt^2 = u_2t - \frac{1}{2}gt^2$

$H = u_2t$

$t = \frac{H}{u_2}$

Given $H = 40$ m and $u_2 = 20$ m/s.

$t = \frac{40}{20} = 2$ seconds.

To find the distance from the ground where they meet, substitute the value of $t$ into the position equation of either particle. Using $y_2(t)$:

$y = y_2(2) = u_2(2) - \frac{1}{2}g(2)^2 = 20(2) - \frac{1}{2}g(4) = 40 - 2g$.

Taking $g = 10 \, m/s^2$:

$y = 40 - 2(10) = 40 - 20 = 20$ meters.